\(\dfrac{n_{C_xH_y}}{n_{O_2}}=\dfrac{1}{10}\) \(\Rightarrow\left\{{}\begin{matrix}n_{C_xH_y}=1\\n_{O_2}=10\end{matrix}\right.\) ( mol )
\(C_xH_y+\left(x+\dfrac{y}{4}\right)O_2\rightarrow xCO_2+\dfrac{y}{2}H_2O\)
1 \(x+\dfrac{y}{4}\) \(x\) \(\dfrac{y}{2}\) ( mol )
Cho Y qua dd H2SO4 đặc `=>` H2O bị hấp thụ
hh khí Z gồm CO2 và O2 dư
\(n_{O_2\left(dư\right)}=10-x-\dfrac{y}{4}\left(mol\right)\)
\(d_{\dfrac{hhk}{H_2}}=19\)
\(\Rightarrow M_{hhk}=38\) \((g/mol)\)
Áp dụng quy tắc đường chéo, ta có:
\(\dfrac{n_{CO_2}}{n_{O_2\left(dư\right)}}=\dfrac{1}{1}\) \(\Rightarrow n_{CO_2}=n_{O_2\left(dư\right)}\)
\(\Leftrightarrow x=10-x-\dfrac{y}{4}\) \(\Leftrightarrow2x+\dfrac{y}{4}=10\) \(\Leftrightarrow8x+y=40\)
Với \(x=4;y=8\) thỏa mãn
`=>` CTPT X: \(C_4H_8\)